# Ray Optics and Optical Instruments - Class 12 - Physics

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## Extra Questions - Ray Optics and Optical Instruments | NCERT | Physics | Class 12

A thin equiconvex lens is made of glass of R.I. 1.5 and its focal length is 0.2. If it acts as a concave lens of $0.5$ m focal length when dipped in a liquid, the R.I. of the liquid is:

A) $\frac{17}{8}$

B) $\frac{15}{8}$

C) $\frac{13}{8}$

D) $\frac{9}{8}$

The correct option is **B) $\frac{15}{8}$**.

The lens makers formula for a thin lens is given by: $$ \frac{1}{f} = (n_l - n_m) \left( \frac{1}{R_1} - \frac{1}{R_2} \right) $$ where:

$f$ is the focal length of the lens,

$n_l$ is the refractive index of the lens material,

$n_m$ is the refractive index of the medium surrounding the lens,

$R_1$ and $R_2$ are the radii of curvature of the lens surfaces.

In air, the focal length $f_1$ of the equiconvex lens (both sides have the same radii) is given to be 0.2 m (200 mm, taking care that it is a convex lens, $f$ is positive), and the refractive index of glass is 1.5, while the refractive index of air is 1. Hence: $$ \frac{1}{f} = (1.5 - 1) \left( \frac{2}{R} \right) $$

When the lens is placed in the liquid, it acts as a concave lens with a focal length of $-0.5$ m (given that it is a concave lens in the liquid, $f$ is negative), meaning: $$ \frac{1}{f_a} = \left(\frac{1.5}{n_{\text{liquid}}} - 1\right) \left( \frac{2}{R} \right) $$

Using the above relations, we can relate the focal lengths $f_1$ and $f_a$: $$ \frac{\frac{1}{f_1}}{\frac{1}{f_a}} = \frac{0.5 - 1}{\left(\frac{1.5}{n_{\text{liquid}}} - 1\right)} $$ $$ \frac{-0.2}{0.5} = \frac{-0.5}{0.8} $$ $$ \frac{0.5 - 1}{1.5/n_{\text{liquid}} - 1} = -2.5 $$ $$ 1.5 - n_{\text{liquid}} = 2.5 (1.5/n_{\text{liquid}} - 1) $$

Solving for $n_{\text{liquid}}$, we get: $$ n_{\text{liquid}} = \frac{15}{8} $$

Thus, the refractive index of the liquid in which the lens behaves as a concave lens with a focal length of $-0.5$ m is **$\frac{15}{8}$**.

The object distance $u$, the image distance $v$, and the magnification $m$ in a lens follow certain linear relations. These are:

(A) $\frac{1}{u}$ versus $\frac{1}{v}$

(B) $m$ versus $u$

(C) $u$ versus $v$

(D) $m$ versus $\frac{1}{u}$

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Which of the following has virtual focus?

A. concave lens

B. convex lens

C. convex mirror

D. both (a) and (c)

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A combination of two thin lenses with focal lengths $f_{1}$ and $f_{2}$ respectively forms an image of a distant object at a distance of $60 \mathrm{~cm}$ when the lenses are in contact. The position of this image shifts by $30 \mathrm{~cm}$ towards the combination when the two lenses are separated by $10 \mathrm{~cm}$. The corresponding values of $f_{1}$ and $f_{2}$ are

A) $30 \mathrm{~cm}, -60 \mathrm{~cm}$

B) $20 \mathrm{~cm}, -30 \mathrm{~cm}$

C) $15 \mathrm{~cm}, -20 \mathrm{~cm}$

D) $12 \mathrm{~cm}, -15 \mathrm{~cm}$

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An image on an object is formed by two plane mirrors. One of the plane mirrors is double the size of the other. In which case will the image be bigger and why? Please check my answer. Ans. The image of the object in both the mirrors would be the same, as it is one of the characteristics of the image formed by a plane mirror. The image formed on the plane mirror would be the exact size of the object.

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If the tube length of an astronomical telescope is $105 \mathrm{~cm}$ long and the magnifying power is 20 for the normal setting, calculate the focal length of the objective.

A) $100 \mathrm{~cm}$

B) $10 \mathrm{~cm}$

C) $20 \mathrm{~cm}$

D) $25 \mathrm{~cm}$

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Match the following names with their corresponding types of telescopes:

Radio | Hubble |

X-ray | GMRT |

Optical | Chandra |